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Full text of "Treatise On Analysis Vol-Ii"


For each x e A, the mapping #f~(^*)(x) = %(x) is continuous on X(A),
by the definition of the weak topology. Moreover, we have

and therefore 9(xy) = (9x)(9y). The relation ye = 1 is obvious. The equality
\<9x\ = p(;c) will follow from (ii) and the definition of p(x) (15.2.7). Consider
therefore the first assertion in (ii). We know already that #(*) 6 Sp(jc) for all
;ce A and x e X(A) (15.3.1). Let us show conversely that, for each /I e Sp(x),
there exists a character x such that x(x) = A, or equivalently that x(x  Xe) = 0.
Since x  he is not invertible, the ideal A(x  he) generated by this element is
distinct from A. In view of (1 5.3.1 (iii)) we are therefore reduced to proving
the following lemma:

( In a separable commutative Banach algebra A with unit element
e 7^ 0, every ideal a ^ A is contained in a maximal ideal.

Let (xn) be a dense sequence in A. We define inductively an increasing
sequence (an) of closed ideals ^A (n ^ 0) as follows. Take a0 == a (which is
distinct from A, by ( In the quotient Banach algebra A/ an.l con-
sider the image xn ofxn and an element Art of its spectrum (which by (15.2.4) is
not empty), so that xn  lne (where e is the canonical image of e in A/an-i)
is not invertible in A/aw-t. It follows that the ideal a'n in A generated by
an_! and xn  ln e is distinct from A, and therefore an = a^ is distinct from A.
The union m' of the ideals an is therefore also an ideal ^ A, and hence
the closure m of m' is also an ideal ^ A. We shall show that m is maximal,
which will complete the proof. If e" is the unit element of A/m, and x"n the
image of xn in A/m, then by construction we have x"n = kne"\ in other words,
all the x belong to the closed subalgebra (5.9.2) Ce" of A" = A/m. Since
they form a dense subset of A" (3.11.4), it follows that A" = O", and there-
fore m is maximal.


(15.3.5) It should be noted that the Gelfand transformation x^&x is
not necessarily injective. For example, if x is nilpotent but not zero, say
of = 0, then the image of Sp(x) under \-> C* is {0} (( and (15.2.1)) and
therefore Sp(x) = {0}, hence %(x) = 0 for all characters %, although x ^ 0.
This example also shows that \\9x\\ need not be equal to ||x||. But even
if ^ is isometric (in which case the image 9(A) is closed'm #C(X(A)) (3.14.4)),
may not be equal to c(*(A)) (15.3.8).nt that